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Question
Select the correct answer from the given alternatives:
f(x) = `{:(= (2^(cotx) - 1)/(pi - 2x)",", "for" x ≠ pi/2),(= log sqrt(2)",", "for" x = pi/2):}`
Options
f is continuous at x = `pi/2`
f has a jump discontinuity at x = `pi/2`
f has a removable discontinuity
`lim_(x -> pi/2) "f"(x)` = 2 log 3
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Solution
f is continuous at x = `pi/2`
Explanation;
`"f"(pi/2) = log sqrt(2)`
`lim_(x -> pi/2) "f"(x) = lim_(x -> pi/2) (2^(cotx) - 1)/(pi - 2x)`
= `lim_(x -> pi/2) (2^(tan(pi/2 - x)) - 1)/(2(pi/2 - x))`
Put `pi/2 - x` = h
As `x -> pi/2, "h" -> 0`
∴ `lim_(x -> pi/2) "f"(x) = lim_("h" -> 0) (2^(tan"h") - 1)/(2"h")`
= `1/2 lim_("h" -> 0) ((2^(tan"h") - 1)/(tan"h") xx tan"h"/"h")` ...[∵ h → 0, ∴ tan h → 0 ∴ tan h ≠ 0)
= `1/2 lim_("h" -> 0) (2^(tan"h") - 1)/tan"h" xx lim_("h" -> 0) tan"h"/"h"`
= `1/2*log 2*(1)`
= `log sqrt(2)`
= `"f"(pi/2)`
∴ f is continuous at x = `pi/2`
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